Precalculus Examples

$\sin (- 15 °)$

Step 1

First, split the angle into two angles where the values of the six trigonometric functions are known. In this case, $- 15 °$ can be split into $45 - 60$ .

$\sin (45 - 60)$

Step 2

Use the difference formula for sine to simplify the expression. The formula states that $\sin (A - B) = \sin (A) \cos (B) - \cos (A) \sin (B)$ .

$\sin (45) \cos (60) - \cos (45) \sin (60)$

Step 3

Remove parentheses.

$\sin (45) \cos (60) - \cos (45) \sin (60)$

Step 4

Simplify each term.

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Step 4.1

The exact value of $\sin (45)$ is $\frac{\sqrt{2}}{2}$ .

$\frac{\sqrt{2}}{2} \cos (60) - \cos (45) \sin (60)$

Step 4.2

The exact value of $\cos (60)$ is $\frac{1}{2}$ .

$\frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \cos (45) \sin (60)$

Step 4.3

Multiply $\frac{\sqrt{2}}{2} \cdot \frac{1}{2}$ .

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Step 4.3.1

Multiply $\frac{\sqrt{2}}{2}$ by $\frac{1}{2}$ .

$\frac{\sqrt{2}}{2 \cdot 2} - \cos (45) \sin (60)$

Step 4.3.2

Multiply $2$ by $2$ .

$\frac{\sqrt{2}}{4} - \cos (45) \sin (60)$

Step 4.4

The exact value of $\cos (45)$ is $\frac{\sqrt{2}}{2}$ .

$\frac{\sqrt{2}}{4} - \frac{\sqrt{2}}{2} \sin (60)$

Step 4.5

The exact value of $\sin (60)$ is $\frac{\sqrt{3}}{2}$ .

$\frac{\sqrt{2}}{4} - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}$

Step 4.6

Multiply $- \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}$ .

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Step 4.6.1

Multiply $\frac{\sqrt{3}}{2}$ by $\frac{\sqrt{2}}{2}$ .

$\frac{\sqrt{2}}{4} - \frac{\sqrt{3} \sqrt{2}}{2 \cdot 2}$

Step 4.6.2

Combine using the product rule for radicals.

$\frac{\sqrt{2}}{4} - \frac{\sqrt{3 \cdot 2}}{2 \cdot 2}$

Step 4.6.3

Multiply $3$ by $2$ .

$\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{2 \cdot 2}$

Step 4.6.4

Multiply $2$ by $2$ .

$\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}$

Step 5

Combine the numerators over the common denominator.

$\frac{\sqrt{2} - \sqrt{6}}{4}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{\sqrt{2} - \sqrt{6}}{4}$

Decimal Form:

$- 0.25881904 \dots$